SolnFinalReview - V33 3‘ u Midis may} ~ " M...

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Unformatted text preview: V33 3‘ u Midis may} ~ " M [WWWJ x a \y, 3“ n W}; 3;} 13 .3” a f N M/ ~. « MW .2 2 3 1,, 0.3. 9 N M ., W 2/ ,, qr. .Mw 1: L um , a M 0 Q4 /, \ 9 a, {. L‘* <1»qu pl“ .9 4.; 4’, ink? 33 fl LfofiFM'é é ; r‘ mm ~é‘ 5x44793<20 “’ LL) gm,“ vwmw»m¢,wm«mummmmw ' \Jx .— \4’2. , “‘31 x E t A” 0 ¢ '\ ‘ 7. A I .I g; 0,937 CH0 M~ WW“, w", 2 w H, a a: In» «9 .... gig]. "WWW? ,. NSF—“Afi » E x Q 71> L3 okv ,T L , IN A ‘7 72 W31 1” ix a, {w 3 EE 411, 2008-02-27 Quiz 6 15650 Name: ____ EID: (Q1) If the switch in the figure below has been closed for a long time before t = O , but is opened at t = 0 determine: (a) the characteristic equation of the circuit, (b) ixand V]: for t >0. (a) Let V = capacitor voltage and i : inductor current. At t = 0-, the switch is closed and the circuit has reached steady—state. v(O‘-) = 16V and i(0—) = 16/8 = 2A At t = 0+, the'switch'is Open but, v(0+) ="16 and i(0+)’ = 2. We now have a source-free RLC circuit. R 8+12 = ZOohms, L 2 1H, C = 4mF. OL = R/(2L) = (20)/(2X1) = 10 (00: 1/\/LC = 1/ 1X(1/36) = 6 Since on > (no, we have a overdarnped response. 81,2 = —oci1/0c2 —co(2, = ~18,-2 Thus, the characteristic equation is (s + 2)(s + 18) = O or 52 + 20s +36 = 0. (b) i(t) = [Ae‘2‘+Be'13‘] and i(O) = 2 = A+B (1) To get di(0)/dt, consider the circuit below at t 2 0+. EE 411, 2003-02-27 Quiz 6 ' 15650 Name: EID: 129 39 + V —V(O) + 201(0) + VL(O) = O, which leads to, ~16 +20x2 +VL(O) = 0 or VL(0) = —24 Ldi(O)/dt = VL(O) which gives di(0)/dt = VL(0)/L = —24/1 = ~24 A/s Hence —24 = —2A— 18B or 12 = A+9B (2) From (1) and (2), B = 1.25 andA = 0.75 1(1) = [0,7552% 1.25e'18t] = —ix(t) or ix(t) = [-0.75e'2t- 1.25e'18t] A Va) = 81(t) = [6e'2‘+ 10e'18‘] A ...
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This note was uploaded on 10/26/2009 for the course EE 411 taught by Professor Lee during the Spring '07 term at University of Texas at Austin.

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SolnFinalReview - V33 3‘ u Midis may} ~ &amp;quot; M...

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